# How does a complex algebraic variety know about its analytic topology?

This question has two parts. The first is a reference request regarding a result I assume is standard, and the second is a soft question asking for philosophy and intuition about an issue the first question raises for me. Apologies in advance that it might be hard to answer completely. Nonetheless I expect and hope that both the question and any answers will be interesting and useful to denizens of the site.

Part 1: the analytic topology is an isomorphism invariant, right?

Let $$V,W$$ be affine algebraic sets (i.e. vanishing sets of ideals of polynomials) embedded in $$\mathbb{C}^m,\mathbb{C}^n$$ respectively, and suppose there exists an isomorphism $$f:V\rightarrow W$$ of algebraic varieties. If $$V,W$$ are equipped with their subspace topologies as subspaces of the Euclidean spaces $$\mathbb{C}^m$$ and $$\mathbb{C}^n$$, is $$f$$ necessarily a homeomorphism?

I assume that the answer is yes and that this is a standard fact, since it seems to me it should be necessary in order to define the analytic topology on a general complex analytic variety. (If the answer is not "yes", then "the analytic topology" is meaningless because it would depend on choices of coordinates.)

On the other hand, it is not obvious to me, because the analytic/Euclidean topology on an algebraic subset of $$\mathbb{C}^n$$ is so much finer than its Zariski topology. How does knowing that $$f$$ is an algebraic isomorphism allow us to control all the extra data in the analytic topology?

This last question suggests the broader inquiry of part 2 to me -

Part 2: how does the algebraic variety know about its analytic topology?

This is the soft question. If two complex algebraic sets can't be isomorphic as varieties without being homeomorphic as subsets of Euclidean space, this suggests that their structure as algebraic varieties "knows something" about this finer topology. In fact, the variety must in some sense "know" the entire analytic topology, as it can be recovered by embedding into affine space, endowing the latter with the Euclidean topology, and taking the subspace topology. Given this -

Where in the data defining the variety (the Zariski topology + the structure sheaf) does the data defining the analytic topology hide?

In other words, it must be that given a subset $$U$$ of a complex algebraic variety $$V$$, one can test whether $$U$$ is open for the analytic topology using only the data contained in $$V$$'s Zariski topology and structure sheaf -

What is the test?

Hoot's comment and KReiser's answer have clarified for me an aspect of Part 2 that I would like to make explicit. As I insinuated at the beginning of part 2, although given a complex algebraic variety (i.e. scheme of finite type over $$\mathbb{C}$$), one can recover the analytic topology by embedding its affines individually in complex affine space and taking the subspace topology, I am somehow unsatisfied by this as an explanation of how the variety knows about its analytic topology. Hoot's comment and KReiser's answer clarify for me why I am unsatisfied: what this is doing, philosophically, is using the sections of the structure sheaf to "carry" the analytic topology on $$\mathbb{C}^n$$ back to the variety, using the fact that the sections can be thought of as functions to $$\mathbb{C}$$. The reason I wasn't satisfied with this (and hence asked the part 2 question) is that the analytic topology on $$\mathbb{C}^n$$ is data that lies outside the data that defines the variety. Hence, I would like to know if there is a way to test $$U\subset V$$ for analytic openness that can query any info contained in the data defining $$V$$ (its Zariski topology and structure sheaf, i.e. the $$\mathbb{C}$$-algebra structure of its rings of sections and the restriction maps) but never queries whether or not a subset of any $$\mathbb{C}^n$$ is analytically open.

Is there a test as above that does not consult the a-priori-known analytic topology on $$\mathbb{C}$$?

Another way to sharpen the question:

KReiser's answer shows that the reason that the analytic topology is an isomorphism invariant is essentially that polynomials are continuous. (KReisner noted even analyticity, so that an algebraic isomorphism is also an analytic isomorphism, but the only feature needed to conclude it's a homeomorphism is continuity.) Now any topology on $$\mathbb{C}$$ induces one on $$\mathbb{C}^n$$, and we can ask with respect to any such topology whether polynomial functions $$\mathbb{C}^n\rightarrow\mathbb{C}^m$$ are continuous. If they are, the exact same proof shows that the given topology is an isomorphism invariant of affine complex algebraic varieties, and thus can be imposed in a well-defined way on an arbitrary variety. The Zariski and analytic topologies are two such topologies. The latter is a refinement of the former.

Can we say anything about the family of topologies described in the last paragraph? How big is it? What do they have in common? Can any of them be recovered from the data defining the variety, without a priori knowledge of the topology? [Obv. the Zariski topology can, because it is part of the data defining the variety.]

• The analytic topology also can be constructed from its algebra of functions(i.e, $X=\mathrm{maxSpec}(\mathcal{O})$ as a set in this case) by defining the basic open sets of $X$ by $f^{-1}(a,b), \forall f\in \mathcal{O}, (a,b)\subseteq \mathbb{R}$. Jul 20, 2016 at 4:24
• My impression is that you want to do this without presupposing that you know anything about the topology of $\mathbb C$ or $\mathbb R$, is that right?
– Hoot
Jul 20, 2016 at 21:01
• @Hoot - ideally - if it's possible! I will add a note to the OP clarifying this. Jul 21, 2016 at 5:05

Answer to part 1: If $f$ is an isomorphism of algebraic varieties, it must be a regular map (which is therefore analytic, as polynomials and their appropriately-taken quotients are analytic). Similarly, it's two-sided inverse must also be regular and thus analytic, and so we have that $f$ is also an isomorphism of analytic varieties. Therefore it must be a homeomorphism.
Attempt at part 2: You've already said the basic response that would jump in to my head first- you can embed a variety into $\mathbb{C}^n$ and give it the subspace topology from the usual analytic topology. In this strategy, the information you're using comes from a choice of generators for $\mathcal{O}_X(X)$- this determines the embedding in to $\mathbb{A}^n_\mathbb{C}$ which then gets you access to the complex topology after taking $\mathbb{C}$-points.
Here's an attempt at a simpler way to test for $U\subset V$ analytically open- if you can build $U$ as a appropriate union and intersection of $f^{-1}(W)$ for some $f$ a section of the Zariski structure sheaf and $W\in \mathbb{C}$ analytically open. This idea could probably use some fleshing out (I'm not completely sure it's correct), but the theme is that regular functions are all ready to work in the analytic world already (as we saw in the answer to part 1).
• +1 Thank you. Part 1 was much easier than I expected. Your answer to part 2 boils down it seems to me to some form of "carrying the analytic topology from $\mathbb{C}$ back to the variety via the functions of the structure sheaf." This and Hoot's comment clarify for me that I'm also curious if it's possible to determine analytic openness of a subset in the variety without accessing info about analytic openness in $\mathbb{C}$. This seems like more of a long shot, but I want to know, so I'll add a note to the OP and leave the question unanswered for a while in hopes of further engagement. Jul 21, 2016 at 5:16